# Properties of holomorphic function having convergent Laurent series in some domain

While trying assignment problems of complex analysis I am unable to solve this particular question and so I am posting it here.

Let $$f$$ be a holomorphic function on $$0<|z|<\epsilon$$ , $$\epsilon >0$$ given by convergent Laurent series $$\sum_{n=-\infty}^{\infty} a_{n} z^{n}$$ . Given also that $$\lim_{z\to0} |f(z)|= \infty$$ .

Then which one is true.

1. $$a_{-1} \neq 0$$ and $$a_{-n} =0$$ for all $$n\ge 2$$.

2. $$a_{-N} \neq 0$$ for some $$N>1$$ and $$a_{-n}=0$$ for all $$n >N$$ .

3. $$a_{-n} =0$$ for all $$n>0$$.

4. $$a_{-n} \neq 0$$ for all $$n>0$$.

Although I have studied Ch- Laurent series from text book Complex variables with applications by Ponnusamy and Silvermanbut I am completely clueless on how this particular problem can be approched.

Can anyone please shed some light.

• Hint: what kind of singularity does $f$ have at $0$? Removable, pole or essential? Commented Jul 31, 2020 at 18:23
• $z^2/f(z)$ is holomorphic on a small disk Commented Jul 31, 2020 at 21:19
• @Robert Israel I think it's essential singularity. But can you please ellaborate on how it might be useful?
– user775699
Commented Aug 4, 2020 at 8:17
• No, $|f|$ can't go to $\infty$ at an essential singularity (see the Casorati-Weierstrass theorem). Commented Aug 4, 2020 at 14:15
• @Robert Israel Statement of Casorati Weirestrauss theorem is that : If f(z) has an essential singularity at z=$z_{0}$ , then f(z) comes arbitrary close to every complex value in each deleted neighborhood of $z_{0}$ . So, I think it can approach $\infty$ .
– user775699
Commented Aug 5, 2020 at 8:17